There are an infinite number of them.
Here are a few:
y = 4
y = 2 + 2
3y = 12
4y + 7 = 5y + 3
y2 - 9y + 20 = 0
y3 + 2y2 -3y + 13 = 5y2 + 3y + 5
Yes it is. The solution is: y=4 .
No. -4 does not equal -2(1).
(4.25, 0.25) is a solution.
The equation (4y = 20) represents a linear equation where (y) is the variable. To solve for (y), divide both sides of the equation by 4, resulting in (y = 5). Thus, the solution to the equation is (y = 5).
Given: x + y = 5 2x + y = 6 We can subtract the second equation from the first to get: -x = -1 x = 1. From there, substituting back in to the first equation yields: 1 + y = 5 y = 4. The solution is (1, 4).
x = 0 and y = 4
Yes it is. The solution is: y=4 .
If for example: 4y = 16 Then: y = 4
No. -4 does not equal -2(1).
(4.25, 0.25) is a solution.
The equation (4y = 20) represents a linear equation where (y) is the variable. To solve for (y), divide both sides of the equation by 4, resulting in (y = 5). Thus, the solution to the equation is (y = 5).
Given: x + y = 5 2x + y = 6 We can subtract the second equation from the first to get: -x = -1 x = 1. From there, substituting back in to the first equation yields: 1 + y = 5 y = 4. The solution is (1, 4).
Possibly the solution to an equation
It is a simultaneous equation and its solution is x = -1 and y = -5
The equation 6 + y = 12 is a simple equation in one variable. If you add 6 to both sides of the equals sign, you get y = 18, and that is the solution.
You have one equation in two unknowns. There is no solution possible without a second (independent) equation in the same two unknowns.
use the formula y-y1=m(x-x1)